Meeting Rooms I — Conflict Detection

Problem Statement

Given an array of meeting time intervals, determine whether a single person can attend all meetings — that is, whether any two meetings overlap.

Input

An array of intervals [[start, end], …] representing meeting times.

[[0,30],[5,10],[15,20]]
[[7,10],[2,4]]

Output

A boolean: true if the person can attend every meeting, false if any two meetings conflict.

false
true

Constraints

• 0 ≤ intervals.length ≤ 10⁴
• intervals[i].length == 2
• 0 ≤ startᵢ < endᵢ ≤ 10⁶

Intuition

Sort by start time. After sorting, any overlapping meetings must sit next to each other. Walk through consecutive pairs — if one starts before the previous one ends, there’s a conflict.

Before sort:  [0,30] [5,10] [15,20]
After sort:   [0,30] [5,10] [15,20]
               └─────┘
               5 < 30 → conflict detected immediately

Solution

def canAttendMeetings(intervals: list[list[int]]) -> bool:
    intervals.sort(key=lambda x: x[0])
 
    for i in range(1, len(intervals)):
        if intervals[i][0] < intervals[i - 1][1]:
            return False
 
    return True

function canAttendMeetings(intervals: number[][]): boolean {
    intervals.sort((a, b) => a[0] - b[0]);
 
    for (let i = 1; i < intervals.length; i++) {
        if (intervals[i][0] < intervals[i - 1][1]) {
            return false;
        }
    }
 
    return true;
}

Complexity

Edge Case Thinking